Physics 214 Solution Set 2 Winter 2017

1. [Jackson, problem 11.10]

(a) For the Lorentz boost and rotation matrices K and S show that

(ǫ · S)3 = −ǫ · S , (1)

(ǫ′ ·K)3 = ǫ′ ·K ,

where ǫ and ǫ′ are any real unit 3-vectors.

We are given

S1 =

0 0 0 00 0 0 00 0 0 −10 0 1 0

, S2 =

0 0 0 00 0 0 10 0 0 00 −1 0 0

, S3 =

0 0 0 00 0 −1 00 1 0 00 0 0 0

,

K1 =

0 1 0 01 0 0 00 0 0 00 0 0 0

, K2 =

0 0 1 00 0 0 01 0 0 00 0 0 0

, K3 =

0 0 0 10 0 0 00 0 0 01 0 0 0

.

To prove eq. (1), we note that the first row and column of S1, S2 and S3 are all zeros. Hencewe can simply focus on the remaining 3× 3 block. That is, we write

(Si)jk =

0 0k

0T

j −ǫijk

,

where 0 is a row vector of three zeros, 0T is a column vector of three zeros, and

ǫijk =

+1 , if (ijk) is an even permutation of (123) ,

−1 , if (ijk) is an odd permutation of (123) ,

0 , otherwise ,

is the three-dimensional Levi-Civita tensor. After excluding the first row and column, jk labelsthe three remaining rows and columns of Si.

Thus, we can compute (ǫ ·S)3 by pretending that the first row and column do not exist. Indetail,1

(ǫ · S)3jk = (ǫ · S)jℓ(ǫ · S)ℓm(ǫ · S)mk = ǫi(Si)jℓ ǫp(Sp)ℓm ǫq(Sq)mk

= −ǫiǫpǫqǫijℓ ǫpℓm ǫqmk = ǫiǫpǫqǫijℓ ǫpmℓ ǫqmk

= ǫiǫpǫq(δipδjm − δimδjp)ǫqmk = ǫqǫqjk − ǫmǫjǫqǫqmk , (2)

1In eq. (2), we employ the Einstein summation convention. In this derivation, we make use of the antisym-metry properties of the Levi-Civita tensor and employ the identity ǫijℓǫpmℓ = δipδjm − δimδjp.

1

after noting that ǫiǫi = ǫ · ǫ = 1 since ǫ is an arbitrary real unit vector. We now observethat ǫmǫjǫqǫqmk = 0 since ǫmǫq is symmetric under the interchange of m and q whereas ǫqmk isantisymmetric under the same interchange of indices. Thus, eq. (2) yields

(ǫ · S)3jk = ǫqǫqjk = −ǫq(Sq)jk = −(ǫ · S)jk ,

which establishes eq. (1). Note that we could have achieved the same result by writing outexplicitly

ǫ · S =

0 0 0 00 0 −ǫ3 ǫ20 ǫ3 0 −ǫ10 −ǫ2 ǫ1 0

,

and then computing (ǫ · S)3 via matrix multiplication. Indeed,

(ǫ · S)2 =

0 0 0 00 −ǫ22 − ǫ23 ǫ1ǫ2 ǫ1ǫ30 ǫ1ǫ2 −ǫ21 − ǫ23 ǫ2ǫ30 ǫ1ǫ3 ǫ2ǫ3 −ǫ21 − ǫ22

,

and

(ǫ · S)3 = (ǫ · S)2ǫ · S = −(ǫ21 + ǫ22 + ǫ23)

0 0 0 00 0 −ǫ3 ǫ20 ǫ3 0 −ǫ10 −ǫ2 ǫ1 0

= −ǫ · S ,

after using the fact that ǫ is a real unit 3-vector.To establish eq. (2), we can follow either of the above two methods. Employing the first

method above, we make use of the block-matrix form for ǫ ′·K

(ǫ ′·K)jk =

0 ǫ′k

ǫ′j 0jk

, (3)

where 0jk stands for the matrix elements of the 3 × 3 zero matrix. In particular, j labels therow and k labels the column. Then,

(ǫ ′·K)3jk =

0 ǫ′ℓ

ǫ′j 0jℓ

0 ǫ′i

ǫ′ℓ 0ℓi

0 ǫ′k

ǫ′i 0ik

=

ǫ ′· ǫ ′ 0i

0T

j ǫ′jǫ′i

0 ǫ′k

ǫ′i 0ik

=

1 0i

0T

j ǫ′jǫ′i

0 ǫ′k

ǫ′i 0ik

=

0 ǫ′k

ǫ′j ǫ′· ǫ ′ 0jk

=

0 ǫ′k

ǫ′j 0jk

= (ǫ ′

·K)jk ,

after using the fact that ǫ′ is is real unit vector. Thus eq. (2) is established. Alternatively, wecan write out the explicit form for the matrix,

ǫ ′·K =

0 ǫ′1 ǫ′2 ǫ′3ǫ′1 0 0 0ǫ′2 0 0 0ǫ′3 0 0 0

,

2

and then compute (ǫ ′·K)3 via matrix multiplication. Indeed,

(ǫ ′·K)2 =

ǫ′ 21 + ǫ′ 22 + ǫ′ 23 0 0 00 ǫ′ 21 ǫ′1ǫ

′2 ǫ′1ǫ

′3

0 ǫ′1ǫ′2 ǫ′ 22 ǫ′2ǫ

′3

0 ǫ′1ǫ′3 ǫ′2ǫ

′3 ǫ′ 23

,

and

(ǫ ′·K)3 = (ǫ ′

·K)2 ǫ ′·K = (ǫ′ 21 + ǫ′ 22 + ǫ′ 23 )

0 ǫ′1 ǫ′2 ǫ′3ǫ′1 0 0 0ǫ′2 0 0 0ǫ′3 0 0 0

= ǫ ′

·K ,

after using the fact that ǫ ′ is a real unit 3-vector.

(b) Use the result of part (a) to show that:

exp(−ζ β ·K

)= I − β ·K sinh ζ + (β ·K)2 [cosh ζ − 1] ,

where I is the 4× 4 identity matrix.

We employ the series expansion for the exponential (which defines the matrix exponential),

exp(−ζ β ·K

)=

∞∑

n=0

(−ζ)nn!

(β ·K)n . (4)

In part (a), we established the following result: (β ·K)3 = β ·K. Hence, it follows that

(β ·K)2n = (β ·K)2 , (β ·K)2n+1 = β ·K , for n = 1, 2, 3, . . . .

Thus, we can rewrite the series given in eq. (4) as

exp(−ζ β ·K

)= I − β ·K

∑

n oddn≥1

ζn

n!+ (β ·K)2

∑

n evenn≥2

ζn

n!, (5)

after using the fact that (β ·K)0 = I is the 4× 4 identity matrix. Using,

∞∑

n=0

ζ2n+1

(2n+ 1)!= sinh ζ ,

∞∑

n=0

ζ2n

(2n)!= cosh ζ ,

and noting that the last summation in eq. (5) starts at n = 2, we end up with

exp(−ζ β ·K

)= I − β ·K sinh ζ + (β ·K)2 [cosh ζ − 1] , (6)

which is the desired result.

3

REMARKS:

To understand the significance of eq. (6), let us write it explicitly in matrix form. It isconvenient to use the block matrix form of eq. (3), where j labels the row and k labels thecolumn,

I =

1 0k

0T

j δjk

, (β ·K)jk =

0 βk

βj 0jk

, (β ·K)2jk =

1 0k

0T

j βjβk

.

(7)Then, eq. (6) yields

exp(−ζ β ·K

)=

cosh ζ −βk sinh ζ

−βj sinh ζ δjk + βjβk(cosh ζ − 1)

.

In class, we identified ζ = tanh−1 β as the rapidity, which satisfies

γ =1√

1− β2= cosh ζ , βγ = sinh ζ .

Hence, after writing ~β = β β = (β1 , β2 , β3), it follows that

exp(−ζ β ·K

)=

γ −γβk

−γβj δjk + (γ − 1)βjβkβ2

,

which we recognize as the boost matrix defined in eq. (11.98) of Jackson.

2. [Jackson, problem 11.13] An infinitely long straight wire of negligible cross-sectional areais at rest and has a uniform linear charge density q0 in the inertial frame K ′. The frame K ′

(and the wire) move with velocity ~v parallel to the direction of the wire with respect to thelaboratory frame K.

(a) Write down the electric and magnetic fields in cylindrical coordinates in the rest frameof the wire. Using the Lorentz transformation properties of the fields, find the components ofthe electric and magnetic fields in the laboratory.

In the rest frame of the wire (i.e. frame K ′), choose the z-axis to point along the wire. Then,to compute the electric field, we draw a cylinder of length L and radius r′, whose symmetryaxis coincides with the z-axis. Applying Gauss’ law in gaussian units,

∮

S

~E ′· n da = 4πQ , (8)

4

where Q is the total charge enclosed inside the cylinder. In cylindrical coordinates (r′, φ′, z′),2

the symmetry of the problem implies that ~E ′(r′, φ′, z′) = E ′(r′)ρ, where E ′(r′) depends onlyon the radial distance from the symmetry axis. Choosing the surface S to be the surface of thecylinder, we have n = r, and so eq. (8) reduces to

2πρ′LE ′(r′) = 4πQ .

Defining the linear charge density (i.e. charge per unit length) by q0 = Q/L, we conclude that3

~E ′(r′) =2q0r′

r . (9)

Since there are no moving charges in the rest frame of the wire, it follows that ~B ′ = 0 .The transformation laws for the electric and magnetic field between reference frames K and

K ′ are given by4

~E = γ[~E ′ − ~β × ~B ′

]− γ2

γ + 1~β(~β · ~E ′) ,

~B = γ[~B ′ + ~β × ~E ′

]− γ2

γ + 1~β(~β · ~B ′) .

For this problem, ~β = β z. Using the results of part (a), and noting that r = r′ (since theradial direction is perpendicular to the direction of the velocity of frame K ′ with respect to K),it follows that

~E =2γq0r

r , ~B =2γβq0r

φ , (10)

where we have used z · r = 0 and z × r = φ.

(b) What are the charge and current densities associated with the wire in its rest frame? Inthe laboratory?

In reference frame K there are no moving charges, so that ~J ′ = 0. The corresponding chargedensity is

ρ′(r′) =q02πr′

δ(r′) . (11)

To check this, let us integrate over a cylinder of length L and arbitrary nonzero radius, whosesymmetry axis coincides with the z-axis. Then,

∫ρ′(r′) dV =

∫ρ′(r′) r′ dr′ dφ dz′ = q0

∫dr′δ(r′)dz′ = q0L = Q .

2We denote the radial coordinate of cylindrical coordinates in frame K ′ to be r′ rather than the moretraditional ρ′, since we reserve the letter ρ for charge density.

3The direction of the unit vectors r, φ and z are the same in framesK and K ′, so no extra primed-superscriptis required on these quantities.

4Eq. (11.149) of Jackson provides the equations to transform the fields from reference frame K to reference

frame K ′. To transform the fields from K ′ to K, simply change the sign of ~β.

5

Since Jµ = (cρ ; ~J) is a four-vector, the relevant transformation law between frames K and K ′

are:

cρ = γ(cρ′ + ~β · ~J ′) , (12)

~J = ~J ′ +γ − 1

β2(~β · ~J ′)~β + γ~βcρ′ . (13)

Plugging in ~J ′ = 0 and the result of eq. (11), and noting that ~β = β z and r′ = r, it followsthat5

ρ(r) =γq02πr

δ(r) , ~J =γβc q02πr

z δ(r) = ρ(r)vz = ρ(r)~v , (14)

after using v ≡ β c.

(c) From the laboratory charge and current densities, calculate directly the electric andmagnetic fields in the laboratory. Compare with the results of part (a).

This is an electrostatics and magnetostatics problem, so we can use Gauss’ law to compute~E and Ampere’s law to compute ~B. The computation of ~E is identical to the one given inpart (a) with q0 replaced by γq0. Hence, it immediately follows from eq. (9) that

~E(r) =2γq0r

r ,

in agreement with eq. (10). Ampere’s law in gaussian units is

∮

C

~B · d~ℓ =4πI

c,

where I is the current enclosed in the loop C. With ~J given by eq. (14),

I =

∫

A

~J · n da =

∫ρ(r)v r dr dφ = γq0v ,

after noting that n = z points along the direction of the current flow and da = rdrdφ isthe infinitesimal area element perpendicular to the current flow. Using the symmetry of theproblem, ~B = B(r)φ. Thus, evaluating Ampere’s law with a contour C given by a circle

centered at r = 0 that lies in a plane perpendicular to the current flow, d~ℓ = rdφ~φ and

2πrB(r) =4πI

c=

4πγq0v

c,

which yields

~B(r) =2γβq0v

rφ ,

after using v = βc, in agreement with eq. (10).

5We can interpret q ≡ γq0 as the linear charge density as observed in reference frame K. This is notunexpected due to the phenomenon of length contraction.

6

3. [Jackson, problem 11.15] In a certain reference frame, a static uniform electric field E0 isparallel to the x-axis, and a static uniform magnetic field B0 = 2E0 lies in the x–y plane,making an angle θ with respect to the x-axis. Determine the relative velocity of a referenceframe in which the electric and magnetic fields are parallel. What are the fields in this framefor θ ≪ 1 and θ → 1

2π?

In frame K, we have~E = E0 x , ~B = Bx x+By y , (15)

with~E · ~B = |~E| | ~B| cos θ = E0B0 cos θ = 2E2

0 cos θ , (16)

after writing |~E| = E0 and | ~B| = B0 = 2E0. It follows that

Bx = 2E0 cos θ , By = 2E0 sin θ . (17)

The reference frame K ′, in which the electric and magnetic fields are parallel, is moving ata velocity ~v ≡ c~β with respect to reference frame K. That is,

~E ′× ~B ′ = 0 . (18)

The electric and magnetic fields in frame K ′ are related to the corresponding fields in frame Kby eq. (11.149) of Jackson. These relations can be rewritten in the following form,

~E ′‖ = ~E‖ , ~B ′

‖ = ~B‖ , (19)

~E ′⊥ = γ

(~E⊥ + ~β × ~B⊥

), ~B ′

⊥ = γ(~B⊥ − ~β × ~E⊥

). (20)

In our notation above, fields with a ‖ subscript are parallel to ~β and fields with a ⊥ subscript

are perpendicular to ~β. For example,

~β × ~E‖ = 0 and ~β · ~E⊥ = 0 ,

which implies that

~E‖ =(~β · ~E)~β

β2and ~E⊥ = ~E − (~β · ~E)~β

β2.

The form of eqs. (19) and (20) suggests that the relative velocity ~v should lie in the z-direction. That is,

~β = β z ,

in which case ~E‖ = Ezz and ~B‖ = Bzz. Since Ez = Bz = 0, it follows from eq. (19) thatE ′

z = B′z = 0. Using eq. (20), the transverse fields are given by

E ′x = γ(Ex − βBy) = γE0(1− 2β sin θ) , E ′

y = γ(Ey + βBx) = 2βγE0 cos θ , (21)

B′x = γ(Bx + βEy) = 2γE0 cos θ , B′

y = γ(Ey + βBx) = γE0(2 sin θ − β) , (22)

after using eqs. (15)—(17). Note that eq. (18) implies that

E ′xB

′y −E ′

yB′x = (~E ′

× ~B ′)z = 0 .

7

Inserting the above results into eqs. (21) and (22) yields,

γ2E20(1− 2β sin θ)(2 sin θ − β)− 4βγ2E2

0 cos2 θ = 0 .

Multiplying out the factors above and writing cos2 θ = 1− sin2 θ, the above equation simplifiesto

2β2 sin θ − 5β + 2 sin θ = 0 .

This is a quadratic equation in β which is easily solved. The larger of the two roots is greaterthan 1, which we reject since 0 ≤ β ≤ 1 (i.e., 0 ≤ v ≤ c). The smaller of the two roots isnon-negative and less than 1. Thus, we conclude that

β =v

c=

5−√25− 16 sin2 θ

4 sin θ. (23)

The two limiting cases are easily analyzed. In the case of θ ≪ 1, we can work to first orderin θ. From eq. (23) we find that β ≃ 2

5θ. Since θ ≪ 1 it follows that β ≪ 1, in which case

γ = (1− β2)−1/2 ≃ 1 +O(β2) .

Since we are working to first order in θ, we also must work to first order in β. In particular wecan neglect terms such as βθ. Hence, in this limiting case, eqs. (21) and (22) yield

~E ′ = 12~B ′ = E0(x+ 2βy) , for β ≃ 2

5θ ≪ 1 ,

where we have neglected terms that are second order (or higher) in β. Finally, in the limit ofθ → 1

2π, eq. (23) yields β = 1

2. Then γ = 2/

√3, and eqs. (21) and (22) yield6

~E ′ = 0 , ~B ′ =√3E0y , for θ = 1

2π . (24)

REMARK:

In fact, it is easy to show that eq. (23) implies that 0 ≤ β ≤ 12. If we multiply the numerator

and denominator of eq. (23) by 5 +√25− 16 sin2 θ, we obtain,

β =4 sin θ

5 +√

25− 16 sin2 θ.

Since the polar angle lies in the range 0 ≤ θ ≤ π or equivalently 0 ≤ sin θ ≤ 1, it followsimmediately that β ≥ 0 (where β = 0 corresponds to sin θ = 0 as expected). Finally, it is easyto verify that

4 sin θ

5 +√

25− 16 sin2 θ≤ 1

2. (25)

6Recall that in class, we showed that the quantity FµνFµν = 1

2ǫµναβF

µνFαβ = −4 ~E· ~B is a Lorentz invariant.

This means that if ~E and ~B are perpendicular in one frame, then they must be perpendicular in all frames.Thus, if θ = 1

2π in frame K and θ = 0 in frame K ′, then it must be true that either the electric field or the

magnetic field (or both) vanish in frame K ′, since the only vector that is both perpendicular and parallel to agiven fixed nonzero vector is the zero vector. This is indeed the case here, as can be seen in eq. (24).

8

Since the denominator on the left hand side above is positive, we can rewrite eq. (25) as

4 sin θ ≤ 12

(5 +

√25− 16 sin2 θ

). (26)

This inequality is manifestly true for sin θ = 0. For sin θ > 0, eq. (26) can be rearranged intothe following form

8 sin θ − 5 ≤√

25− 16 sin2 θ . (27)

Squaring both sides and simplifying the resulting expression then yields

sin θ (sin θ − 1) ≤ 0 . (28)

Dividing both sides of the equation by sin θ (which is assumed positive) yields sin θ ≤ 1, whichis valid for all polar angles θ. Hence, eq. (25) is established. The inequality becomes an equalityif sin θ = 1, in which case β = 1

2.

4. [Jackson, problem 11.18] The electric and magnetic fields of a particle of charge q movingin a straight line with speed v = βc, given by eq. (11.52) of Jackson, become more and moreconcentrated as β → 1, as indicated in Fig. 11.9 on p. 561 of Jackson. Choose axes so thatthe charge moves along the z axis in the positive direction, passing the origin at t = 0. Letthe spatial coordinates of the observation point be (x, y, z) and define the transverse vector ~r⊥,with components x and y. Consider the fields and the source in the limit of β = 1.

(a) Show that the fields can be written as

~E = 2q~r⊥

r2⊥δ(ct− z) , ~B = 2q

v × ~r⊥

r2⊥δ(ct− z) , (29)

where v is a unit vector in the direction of the particle’s velocity.

We begin with eq. (11.154) on p. 560 of Jackson,

~E =q ~R

R3γ2(1− β2 sin2 ψ)3/2, (30)

where ψ is the angle between the vectors ~v and ~R. I have modified Jackson’s notation byemploying the symbol ~R for the vector that points from the charge q to the observation point~r = (x, y, z) in reference frame K.7 Eq. (30) was also derived in class along with the corre-sponding result for the magnetic field,

~B =q(~v × ~R)

cR3γ2(1− β2 sin2 ψ)3/2. (31)

7Jackson denotes the vector that points from the charge q to the observation point (x, y, z) by ~r. However,I prefer to employ ~r to represent the vector that points from the origin of reference frame K to the observationpoint, as shown in Fig. 1.

9

q

(x, y, z)

~R

~vt

~r

x

z

ψ

Figure 1: A charge q moving at constant velocity ~v in the z-direction as seen from referenceframe K. The angle ψ is defined so that v · R = cosψ.

The reference frame K is exhibited in Fig. 1. It is evident from this figure that

~R = ~r − ~vt . (32)

The velocity vector is taken to lie along the z-direction. That is, ~v = vz.It is convenient to introduce the notation where

~r⊥ = xx + yy , ~r‖ = z~z ,

so that ~r⊥ · ~v = 0 and ~r‖ × ~v = 0. Likewise, we can resolve the vector ~R into componentsparallel and perpendicular to the velocity vector,

~R = ~R‖ + ~R⊥ ,

where~R‖ ≡ R‖z = (z − vt)z , ~R⊥ = ~r⊥ . (33)

after making use of eq. (32). In particular, note that | ~R⊥| ≡ R⊥ = R sinψ. It follows that

R3(1− β2 sin2 ψ)3/2 = (R2 − β2R2 sin2 ψ)3/2 = (R2⊥ +R2

‖ − β2R2⊥)

3/2

= [R2‖ +R2

⊥(1− β2)]3/2 = (R2‖ +R2

⊥/γ2)3/2 . (34)

Note that in obtaining eq. (34) we used R2 = R2⊥ + R2

‖ and γ ≡ (1 − β2)−1/2. Moreover, since

~R⊥ = ~r⊥ [cf. eq. (33)], we may replace R⊥ with r⊥ ≡ |~r⊥| = (x2+y2)1/2 in the above formulae.Eqs. (30), (33) and (34) then yield

~E =γq

[~r⊥ + (z − vt)z

]

(γ2R2‖ + r2⊥)

3/2. (35)

Likewise, eqs.(31), (33) and (34) yield

~B =γq(~v × ~r⊥)

c(γ2R2‖ + r2⊥)

3/2. (36)

10

Consider the limit of β → 1. In this limit, γ → ∞, and we see that

limγ→∞

γ

(γ2R2‖ + r2⊥)

3/2=

{0 , if R‖ 6= 0 ,

∞ , if R‖ = 0 .

This implies that

limγ→∞

γ

(γ2R2‖ + r2⊥)

3/2= kδ(R‖) , (37)

for some constant k. Note that in light of eq. (33),

limγ→∞

R‖ = z − ct ,

since γ → ∞ in the limit of v → c. To determine k, we integrate eq. (37) from −∞ to ∞,since R‖ can be any real number (either positive, negative or zero) depending on the value ofthe time t. Thus, employing the substitution u = γR‖,

k =

∫ ∞

−∞

γ dR‖

(γ2R2‖ + r2⊥)

3/2=

∫ ∞

−∞

du

(u2 + r2⊥)3/2

=u

r2⊥(u2 + r2⊥)

1/2

∣∣∣∣∞

−∞

=2

r2⊥.

Hence, we conclude that

limγ→∞

γ

(γ2R2‖ + r2⊥)

3/2=

2

r2⊥δ(z − ct) .

Inserting this result back into eqs. (35) and (36) yields

~E = 2q~r⊥

r2⊥δ(z − ct) , ~B = 2q

v × ~r⊥

r2⊥δ(z − ct) . (38)

In obtaining ~E above, we noted that in the limit of v → c, the z-component of the electric fieldis proportional to (z − ct)δ(z − ct) = 0 due to a property of the delta function. In obtaining~B above, we noted that limv→c ~v/c = v. Finally, since the delta function is an even functionof its argument, we can write δ(z − ct) = δ(ct− z), and eq. (29) is verified.

(b) Show by substitution into the Maxwell equations that these fields are consistent withthe 4-vector source density

Jα = qcvαδ(2)(~r⊥)δ(ct− z) ,

where the 4-vector vα = (1 ; v).

The four-vector current is given by Jµ = (cρ ; ~J). Hence, using the Maxwell equations ingaussian units,

~∇· ~E = 4πρ =4πJ0

c.

Hence, using eq. (38) and noting that Ez = 0, it follows that

J0 =c

4π~∇· ~E =

c

4π

(~∇⊥ ·

~E +∂Ez

∂z

)=qc

2πδ(z − ct)~∇⊥ ·

(~r⊥

r2⊥

), (39)

11

where ~∇⊥ ≡ x ∂/∂x + y ∂/∂y. For ~r⊥ ≡ x x+ y y 6= 0, an elementary computation yields

~∇⊥ ·

(~r⊥

r2⊥

)=

∂

∂x

(x

x2 + y2

)+

∂

∂y

(y

x2 + y2

)=

y2 − x2

(x2 + y2)2+

x2 − y2

(x2 + y2)2= 0 . (40)

To determine the behavior at ~r⊥ = 0, we consider the two-dimensional analogue of the diver-gence theorem,

∫

A

dxdy ~∇⊥ ·

(~r⊥

r2⊥

)=

∮

C

r⊥dφ~r⊥

r2⊥·r⊥ =

∫ 2π

0

dφ = 2π , (41)

where A is a circular disk and C is the circular boundary of the disk. Note that r⊥ = ~r⊥/r⊥is the outward normal to the circular boundary.

Eqs. (40) and (41) imply that

~∇⊥ ·

(~r⊥

r2⊥

)= 2πδ(2)(~r⊥) , (42)

where δ(2)(~r⊥) is a two-dimensional delta function. Inserting this result into eq. (39), we endup with

J0 = qcδ(2)(~r⊥)δ(z − ct) . (43)

Next, we employ the Maxwell equation,

~∇× ~B − 1

c

∂ ~E

∂t=

4π

c~J , (44)

to evaluate ~J . First, we compute

v × ~r⊥ = z × (xx+ yy) = xy − yx , (45)

where we have used the fact that ~v points in the z direction. It then follows that

~∇×

[v × ~r⊥

r2⊥δ(z − ct)

]= det

x y z

∂

∂x

∂

∂y

∂

∂z−y

x2 + y2δ(z − ct)

x

x2 + y2δ(z − ct) 0

= −xx + yy

x2 + y2δ′(z − ct) +

{∂

∂x

(x

x2 + y2

)+

∂

∂y

(y

x2 + y2

)}δ(z − ct)

= −~r⊥

r2⊥δ′(z − ct) + z ~∇⊥ ·

(~r⊥

r2⊥

)δ(z − ct)

= −~r⊥

r2⊥δ′(z − ct) + 2πz δ(2)(~r⊥)δ(z − ct) , (46)

12

after making use of eq. (42). The prime refers to differentiation with respect to z. Finally, wecompute

∂

∂t

(~r⊥

r2⊥δ(z − ct)

)= −c ∂

∂z

(~r⊥

r2⊥δ(z − ct)

)= −c~r⊥

r2⊥δ′(z − ct) . (47)

Inserting eq. (38) into eq. (44) and using eqs. (46) and (47), we obtain

~J =qc

2π~∇×

[v × ~r⊥

r2⊥δ(z − ct)

]− q

2π

∂

∂t

(~r⊥

r2⊥δ(z − ct)

)

=qc

2π

{−~r⊥

r2⊥δ′(z − ct) + 2πz δ(2)(~r⊥)δ(z − ct) +

~r⊥

r2⊥δ′(z − ct)

}

= qcv δ(2)(~r⊥)δ(z − ct) , (48)

after using the fact that v = z. Combining eqs. (43) and (48), we can write

Jα = qcvαδ(2)(~r⊥)δ(z − ct) ,

where the four-vector vα = (1 ; v).

(c) Show that the fields of part (a) are derivable from either of the following 4-vectorpotentials:

A0 = Az = −2qδ(ct− z) ln(λr⊥) , ~A⊥ = 0 , (49)

orA0 = Az = 0 , ~A⊥ = −2qΘ(ct− z)~∇⊥ ln(λr⊥) , (50)

where λ is an irrelevant parameter setting the scale of the logarithm. Show that the twopotentials differ by a gauge transformation and find the corresponding gauge function χ.

The four-vector potential is Aµ = (Φ ; ~A). Given the four-vector potential, the electromagneticfields are determined by

~E = −~∇A0 − 1

c

∂ ~A

∂t, ~B = ~∇× ~A .

Inserting the scalar and vector potentials given in eq. (49),

~E = 2q~∇[δ(ct− z) ln(λr⊥)

]+

2q

cz ln(λr⊥)

∂

∂tδ(ct− z)

= 2qδ(z − ct)

(x∂

∂x+ y

∂

∂y

)[12ln(x2 + y2) + lnλ

]+ 2qz ln(λr⊥)

(∂

∂z+

1

c

∂

∂t

)δ(ct− z)

= 2q~r⊥

r2⊥δ(z − ct) ,

after using ~r⊥ = xx+ yy and r2⊥ = x2 + y2. In particular, note that(∂

∂z+

1

c

∂

∂t

)f(ct− z) = 0 ,

13

for any function of ct− z. Using eq. (49) to compute the magnetic field,

~B = ~∇× ~A = det

x y z

∂

∂x

∂

∂y

∂

∂z

0 0 −2q ln(λr⊥)δ(ct− z)

= −2qδ(ct− z)

{x∂

∂yln(λr⊥)− y

∂

∂xln(λr⊥)

}

= −2qδ(ct− z)

{x∂

∂y

[12ln(x2 + y2) + lnλ

]− y

∂

∂x

[12ln(x2 + y2) + lnλ

]}

= −2q

r2⊥(yx− xy)δ(ct− z) = 2q

v × ~r⊥

r2⊥δ(ct− z) ,

after employing eq. (45).Repeating these calculations using eq. (50),

~E = −1

c

∂ ~A⊥

∂t= 2qδ(ct− z)

(x∂

∂x+ y

∂

∂y

)[12ln(x2 + y2) + lnλ

]= 2q

~r⊥

r2⊥δ(ct− z) ,

after using the relation between the delta function and the step function, δ(x) =d

dxΘ(x). In

the computation of the magnetic field, we require the following result:

~∇⊥ ln(λr⊥) =

(x∂

∂x+ y

∂

∂y

)[12ln(x2 + y2) + lnλ

]=xx+ yy

x2 + y2.

Hence, it follows that

~B = ~∇× ~A = −2q det

x y z

∂

∂x

∂

∂y

∂

∂z

Θ(ct− z)x

x2 + y2Θ(ct− z)

y

x2 + y20

=yx− xy

x2 + y2δ(ct− z) + zΘ(ct− z)

{∂

∂x

(y

x2 + y2

)− ∂

∂y

(x

x2 + y2

)}

= 2qv × ~r⊥

r2⊥δ(ct− z) ,

after employing eq. (45) and noting that

∂

∂x

(y

x2 + y2

)− ∂

∂y

(x

x2 + y2

)= − 2xy

(x2 + y2)2+

2xy

(x2 + y2)2= 0 .

14

Finally, we demonstrate that eqs. (49) and (50) differ by a gauge transformation. Under agauge transformation (using gaussian units),

~A → ~A ′ = ~A+ ~∇χ , A0 → A′ 0 = A0 − 1

c

∂χ

∂t.

Denoting Aµ by eq. (49) and A′µ by eq. (50), it follows that

∂χ

∂t= −2qcδ(ct− z) ln(λr⊥) ,

~∇⊥χ = −2qΘ(ct− z)~∇⊥ ln(λr⊥) ,

∂χ

∂z= 2qδ(ct− z) ln(λr⊥) .

The solution to these equations can be determined by inspection,

χ(~x, t) = −2qΘ(ct− z) ln(λr⊥) ,

up to an overall additive constant.

5. [Jackson, problem 11.27]

(a) A charge density ρ′ of zero total charge, but with a dipole moment ~p ′, exists in reference

frame K ′. There is no current density in K ′. The frame K ′ moves with velocity ~v = ~βc in theframe K. Find the charge and current densities ρ and ~J in the frame K and show that thereis a magnetic dipole moment ~m = 1

2(~p ′ × ~β), correct to first order in β. What is the electric

dipole moment ~p in K to the same order in β?

We shall assume that in frame K ′, we have ~J ′ = 0 and the charge density, ρ′(~x ′) is time-independent. Furthermore, we assume that the total charge in frame K ′ is zero, i.e.

∫ρ′(~x ′) d3x′ = 0 , (51)

whereas the electric dipole moment in frame K ′, denoted by ~p ′, is assumed to be nonzero. Bydefinition, the electric dipole moment is given by

~p ′ =

∫~x ′ ρ′(~x ′) d3x′ . (52)

Frame K ′ is assumed to move with velocity ~v = ~βc with respect to frame K. We shall employeqs. (12) and (13), which we repeat here:

cρ = γ(cρ′ + ~β · ~J ′) , (53)

~J = ~J ′ +γ − 1

β2(~β · ~J ′)~β + γ~βcρ′ . (54)

15

Setting ~J ′ = 0, it follows that

ρ(~x) = γρ′(~x ′) , ~J(~x) = γ~βcρ′(~x ′) = ~βcρ(~x) . (55)

Hence, the electric dipole moment in frame K is given by

~p =

∫~x ρ(~x) d3x , (56)

and the magnetic dipole moment is defined (in gaussian units) in frame K as

~m =1

2c

∫~x× ~J(~x) d3x = 1

2

∫~x× ~β ρ(~x) d3x = 1

2

{∫~x ρ(~x) d3x

}× ~β ,

after employing eq. (55) in the second step above. In light of eq. (56), it follows that

~m = 12~p× ~β . (57)

In order to relate ~p to ~p ′, we need to compare ρ(~x) and ρ′(~x ′). Eq. (11.19) of Jackson gives

the four-vector x′µ in terms of xµ. The inverse relation is obtained by changing the sign of ~β,which yields

x0 = γ(x′0 +~β · ~x ′) , ~x = ~x ′ +

γ − 1

β2(~β · ~x ′)~β + γ~βx′0 . (58)

Working to first order in β, we can set γ = (1− β2)−1/2 ≃ 1, and

x0 ≃ x′0 + ~β · ~x ′ , ~x ≃ ~x ′+ ~βx′0 . (59)

In the same approximation, d3x′ ≃ d3x and ρ(~x) ≃ ρ′(~x ′). Hence, using eq. (51),

~p =

∫~x ρ(~x) d3x ≃

∫(~x ′+ ~βx′0)ρ

′(~x ′) d3x′ =

∫~x ′ρ′(~x ′) d3x′ = ~p ′ . (60)

to first order in β. Using eqs. (57) and (60), we conclude that to first order in β,

~m = 12~p ′ × ~β .

(b) Instead of the charge density, but no current density, in K ′, consider no charge density,

but a current density ~J ′ that has a magnetic dipole moment ~m′. Find the charge and currentdensities in K and show that to first order in β there is an electric dipole moment ~p = ~β× ~m ′

in addition to the magnetic dipole moment.

Suppose that ρ′(~x ′) = 0 in frame K ′. Then, using the continuity equation

~∇ · ~J ′ +∂ρ′

∂t= 0 , (61)

16

and it follows that ~J ′ is a steady current; that is,

~∇ ′· ~J ′(~x ′) = 0 . (62)

The magnetic dipole moment in frame K ′ is nonzero and is denoted by ~m ′. By definition, themagnetic dipole moment is given by

~m′ =1

2c

∫~x ′

× ~J ′(~x ′) d3x′ . (63)

Then, using eqs. (53) and (54),

ρ(~x) =γ

c~β · ~J ′(~x ′) , (64)

~J(~x) = ~J ′(~x ′) +γ − 1

β2[~β · ~J ′(~x ′)] ~β . (65)

Taking the dot product of eq. (65) with ~β,

~β · ~J(~x) = γ~β · ~J ′(~x ′) . (66)

Inserting this result into eq. (64) yields

ρ(~x) =~β

c· ~J(~x) . (67)

Using eq. (67), the electric dipole moment in frame K is given by

~p =

∫~x ρ(~x) d3x =

1

c

∫~x(~β · ~J(~x)

)d3x . (68)

Working to first order in β, we can set γ = (1 − β2)−1/2 ≃ 1, in which case d3x′ ≃ d3x and~β · ~J(~x) = ~β · ~J ′(~x ′). Hence,

~p ≃ 1

c

∫~x(~β · ~J ′(~x ′)

)d3x′ .

We now apply using eq. (59) which yields,

~p ≃ 1

c

∫~x ′

(~β · ~J ′(~x ′)

)d3x′ +

~β x′0c

~β ·

∫~J ′(~x ′) d3x′ . (69)

This result can be simplified by using the following trick. One can express the ith componentof the current as follows,

J ′ i = ∇′k(J

′ kx′ i)− x′ i ~∇′· ~J ′ = ∇′

k(J′ kx′ i) , (70)

after applying eq. (62) in the final step. Thus,∫J ′ i d3x′ =

∫∇′

k(J′ kx′ i) d3x′ = 0 , (71)

17

after converting to a surface integral at infinity, which vanishes under the assumption of alocalized current. Hence, the last term in eq. (69) vanishes, and we are left with

~p ≃ 1

c

∫~x ′

(~β · ~J ′(~x ′)

)d3x′ . (72)

We can rewrite the integrand of eq. (72) by using the vector identity

~x ′(~β · ~J ′(~x ′)

)= ~β ×

(~x ′

× ~J ′(~x ′))+ ~J ′(~x ′)(~β · ~x ′) .

Inserting this result into eq. (72) and using eq. (63), it follows that

~p = 2~β × ~m ′ +1

c

∫~J ′(~x ′)(~β · ~x ′) d3x′ . (73)

The remaining integral can evaluated by employing eq. (70) for ~J ′. In particular, the ithcomponent of the integral in eq. (73) is

1

c

∫J ′ iβjx′ j d3x′ =

βj

c

∫x′ j ∇′

k(J′ kx′ i) d3x′ = −β

j

c

∫J ′ kx′ i∇′

k(x′ j) d3x′ = −β

j

c

∫J ′ jx′ i d3x′ ,

(74)after an integration by parts, where the surface integral at infinity can be dropped as it vanishesfor a localized current. In the final step of eq. (74), we used ∇′

k(x′ j) = δjk and summed over

the repeated index k. Thus, eq. (74) yields

∫~J ′(~x ′)(~β · ~x ′) d3x′ = −1

c

∫~x ′

(~β · ~J ′(~x ′

)d3x′ ≃ −~p ,

where we have used eq. (72) in the final step. Inserting this result back into eq. (73) yields

~p = 2~β × ~m ′ − ~p .

Solving for ~p, we end up with~p = ~β × ~m ′ . (75)

For completeness, we shall evaluate ~m in terms of ~m′, keeping only terms to first orderin β. In this approximation, we have d3x ≃ d3x′ and ~J(~x) = ~J ′(~x ′). Using these results alongwith eq. (59), it follows that

~m =1

2c

∫~x× ~J(~x) d3x =

1

2c

∫~x× ~J ′(~x ′) d3x′ =

1

2c

∫(~x ′ + ~βx′0)× ~J ′(~x ′) d3x′

= ~m′ +x′02c

~β ×

∫~J ′(~x ′) d3x′ ,

after using eq. (63). However, in light of eq. (71), the remaining integral above vanishes. Hence,to first order in β,

~m = ~m′ . (76)

18

ADDENDUM to Problem 5

Although Jackson only asks for the the results to first order in β, it is not too hard to derivethe exact results. This derivation will be given below.

(a) By assumption, ρ′(~x ′) is a time-independent charge density in frame K ′ which satisfieseq. (51) [i.e. the total charge vanishes]. We are asked to compute ~p and ~m in frame K. Byassumption, this means that we should evaluate these quantities at a fixed time x0 in frame K.Using eq. (58), it follows that

x′0 =x0γ

− ~β · ~x ′ .

Inserting this result back into eq. (58) then yields

~x = ~x ′ +

[γ − 1

β2− γ

](~β · ~x ′)~β + ~βx0 .

We can simplify this expression by noting that γ = (1 − β2)−1/2 implies that β2γ2 = γ2 − 1.Substituting for β2 above,

γ − 1

β2− γ = − γ

γ + 1. (77)

Hence,

~x = ~x ′ − γ

γ + 1(~β · ~x ′)~β + ~βx0 . (78)

We can evaluate the Jacobian of the transformation, eq. (78),8

∂xi∂x′j

= δij −γ

γ + 1βiβj , (79)

which we evaluate at fixed x0. To compute the determinant of the Jacobian, we use the followinggeneral result which is easily proved (see the Appendix following this addendum):

det(δij − aiaj) = 1− |~a|2 .

Hence,

det

(∂xi∂x′j

)= 1− γβ2

γ + 1=

1

γ + 1

[γ + 1− γ2 − 1

γ

]=

1

γ.

It follows that

d3x = det

(∂xi∂x′j

)d3x′ =

d3x′

γ, (80)

which is just the well-known length contraction in special relativity. Using eqs. (55) and (80),it follows that the charge dq located inside an infinitesimal volume element is

dq = ρ(~x) d3x = ρ′(~x ′) d3x′ , (81)

8In eq. (79), we do not distinguish between lowered and raised indices, as all involved quantities are three-dimensional.

19

since the factors of γ cancel out. Eq. (81) is expected since the electric charge is a Lorentzscalar, which must be independent of the reference frame used to evaluate it.

We now can compute ~p in frame K. Using eqs. (78) and (81),

~p =

∫~x ρ(~x) d3x =

∫~x ρ′(~x ′) d3x′ =

∫ [~x ′ − γ

γ + 1(~β · ~x ′)~β + ~βx0

]ρ′(~x ′) d3x′ . (82)

Using eq. (52), it follows that

~p = ~p ′ − γ

γ + 1(~β · ~p ′)~β (83)

where we have used eq. (51) [i.e., the total charge in frame K ′ vanishes] to eliminate the lastterm in eq. (82). Indeed, if we work to first order in β, then the second term on the right handside of eq. (83) can be dropped and we recover the result of eq. (60).

To obtain ~m, we start with the exact result obtained in eq. (57). Inserting eq. (83) for ~p

and using ~β × ~β = 0, it follows that9

~m = 12~p ′ × ~β , (84)

which is exact to all orders in β.

REMARK:

One can derive eq. (83) more directly as follows. Using eqs. (55) and (58),

~p =

∫~x ρ(~x) d3x = γ

∫ [~x ′ +

γ − 1

β2(~β · ~x ′)~β + γ~βx′0

]ρ′(~x ′) d3x

= γ

∫ [~x ′ +

γ − 1

β2(~β · ~x ′)~β + γ~βx′0

]ρ′(~x ′) δ(x0 − ct)d4x .

In the last step, I inserted the integral∫δ(x0 − ct)dx0 = 1 which does not change the result.

This is useful, since d4x = d4x′ = d3x′dx′0. Using eq. (58) to express x0 in terms of x′0,

~p =

∫ [~x ′ +

γ − 1

β2(~β · ~x ′)~β + γ~βx′0

]ρ′(~x ′) δ

(x′0 +

~β·~x ′ − ct

γ

)d3x′dx′0 ,

after writing γδ(x0 − ct) = δ([x0 − ct]/γ) = δ(x′0 +~β·~x ′ − ct/γ). Integrating over x′0 yields

~p =

∫ [~x ′ +

(γ − 1

β2− γ

)(~β · ~x ′)~β

]ρ′(~x ′) d3x′ + ct~β

∫ρ′(~x ′) d3x′ .

Using eqs. (51) and (77) along with ~p ′ =∫~x ′ρ′(~x ′) d3x′, we end up with

~p = ~p ′ − γ

γ + 1(~β · ~p ′)~β ,

which reproduces eq. (83). Of course, the two derivations are equivalent.

9In the literature, one often finds the result of eq. (84) quoted without the factor of 1

2. For a discussion

of this discrepancy, see V. Hnizdo, Magnetic dipole moment of a moving electric dipole, American Journal ofPhysics 80, 645–647 (2012).

20

(b) By assumption, ~J ′(~x ′) is a time-independent, steady current in frame K ′ which satisfieseq. (62). We are asked to compute ~p and ~m in frame K. In this computation, it will be quiteuseful to derive a formula for the following integral,

∫x′ iJ ′j d3x′ .

This can be accomplished using a identity similar to that of eq. (70). First we write

x′ iJ ′ j = 12(x′ iJ ′j + x′ jJ ′ i) + 1

2(x′ iJ ′ j − x′ jJ ′ i) .

We then observe that

x′ iJ ′ j + x′ jJ ′ i = ∇′k(x

′ ix′ jJ ′ k) + x′ ix′ j ~∇′· ~J ′ = ∇′

k(x′ ix′ jJ ′ k) ,

after applying eq. (62), and

x′ iJ ′j − x′ jJ ′ i = ǫijk(~x ′× ~J ′)k .

Consequently,∫x′ iJ ′j d3x′ = 1

2

∫(x′ iJ ′ j + x′ jJ ′ i) d3x′ + 1

2

∫(x′ iJ ′ j − x′ jJ ′ i) d3x′

= 12

∫∇′

k(x′ ix′ jJ ′ k) d3x′ + 1

2ǫijk

∫(~x ′

× ~J ′)k d3x′ . (85)

The first term in the second line of eq. (85) can be converted into a surface integral at infinity,which vanishes under the assumption that the current is localized. The second term is relatedto the magnetic moment via eq. (63). Hence,

1

c

∫x′ iJ ′j d3x′ = ǫijkm′ k . (86)

We are now ready to compute ~m. We again perform the computation at fixed x0, in whichcase we can use eqs. (78) and (80). Employing eq. (65),

~m =1

2c

∫~x× ~J(~x) d3x

=1

2γc

∫ [~x ′ − γ

γ + 1(~β · ~x ′)~β + ~βx0

]×

{~J ′(~x ′) +

γ − 1

β2[~β · ~J ′(~x ′)] ~β

}d3x′

=~m ′

γ+

1

2γc

∫ {γ − 1

β2[~β · ~J ′(~x ′)]~x ′

× ~β − γ

γ + 1(~β · ~x ′)~β × ~J ′(~x ′) + x0~β × ~J ′(~x ′)

}d3x′ ,

where some of the cross-terms vanish due to ~β × ~β = 0. It is convenient to rewrite the aboveexpression in component form,

mk =m′ k

γ+

1

2γcǫijk

∫ {γ − 1

β2x′ iJ ′ ℓβjβℓ − γ

γ + 1x′ ℓJ ′ jβiβℓ + x0J

′ jβi

}d3x′ .

21

Using eqs. (71) and (86), the integrals above are easily evaluated,

mk =m′ k

γ+

1

2γǫijk

{γ − 1

β2ǫiℓnm′nβjβℓ − γ

γ + 1ǫℓjnm′nβiβℓ

}.

Summing over the product of ǫ-tensors,

mk =m′ k

γ+

1

2γm′ n

{γ − 1

β2(δjℓδkn − δjnδkℓ)βjβℓ − γ

γ + 1(δiℓδkn − δinδℓk)βiβℓ

}

=m′ k

γ+

1

2γm′ n(β2δkn − βkβn)

[γ − 1

β2− γ

γ + 1

]

=m′ k

γ+

γ − 1

2(γ + 1)

[β2m′ k − (~β · ~m ′)βk

],

after using β2 = (γ2 − 1)/γ2. We can also use the latter to combine the first two terms above,

1

γ+β2(γ − 1)

2(γ + 1)=

1

2

(1 +

1

γ2

).

Thus, we end up with

~m =1

2

(1 +

1

γ2

)~m ′ − γ − 1

2(γ + 1)(~β · ~m ′)~β .

To first order in β, we can approximate γ ≃ 1, and it follows that ~m ≃ ~m ′, which confirms theresult of eq. (76).

Next, we evaluate ~p at fixed x0 starting from eq. (68),

~p =1

c

∫~x(~β · ~J(~x)

)d3x =

1

c

∫~x(~β · ~J ′(~x ′)

)d3x′ ,

after using eqs. (66) and (80). We now employ eq. (78) and obtain

~p =1

c

∫ {~x ′ − γ

γ + 1~β (~β · ~x ′) + x0~β

} (~β · ~J ′(~x ′)

)d3x′ ,

which we rewrite in component form,

pi =1

c

∫ {x′ i − γ

γ + 1βiβkx′ k + x0β

i

}βjJ ′ j d3x′ .

Using eqs. (71) and (86), the integrals above are easily evaluated,

pi = βj

{ǫijℓm′ ℓ − γ

γ + 1βiβkǫkjℓm′ ℓ

}= (~β × ~m ′)k ,

after noting that ǫkjℓβjβk = 0. Hence,

~p = ~β × ~m ′ ,

which is exact to all orders in β.

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REMARK:

The results of parts (a) and (b) can be combined by linear superposition into the followingresult. Suppose a time-independent localized charge density ρ′(~x ′) of zero total charge and a

time-independent, steady localized current density ~J ′(~x ′) exist in a reference frame K ′. In thisreference frame, the system of charges and currents possesses an electric dipole moment ~p ′ anda magnetic dipole moment ~m ′. The frame K ′ moves with velocity ~v = ~βc in the frame K.Then, in frame K there is a charge density and current density given by:

cρ(~x) = γ(cρ′(~x ′) + ~β · ~J ′(~x ′)

), ~J(~x) = ~J ′(~x ′) +

γ − 1

β2

(~β · ~J ′(~x ′)

)~β + γ~βcρ′(~x ′) ,

where γ = (1− β2)−1/2. Moreover, the electric and magnetic dipole moments in frame K are:

~p = ~p ′ − γ

γ + 1(~β · ~p ′) ~β + ~β × ~m ′ ,

~m =1

2

(1 +

1

γ2

)~m ′ − γ − 1

2(γ + 1)(~β · ~m ′) ~β + 1

2~p ′

× ~β .

The above results are exact to all orders in β. If one sets ~m ′ = 0 [or ~p ′ = 0], then one recoversthe results of part (a) [part (b)] above, respectively.

An equivalent result can be found by defining the unit vector β ≡ ~β/β. Then, usingβ2 = (γ2 − 1)/γ2, we end up with10

~p = ~p ′ −(1− 1

γ

)(β · ~p ′) β + ~β × ~m ′ , (87)

~m =1

2

{(1 +

1

γ2

)~m ′ −

(1− 1

γ

)2

(β · ~m ′) β + ~p ′ × ~β

}. (88)

10Eqs. (87) and (88) were obtained in George P. Fisher, The electric dipole moment of a moving magnetic

dipole, American Journal of Physics 39, 1528–1533 (1971). See also Marijan Ribaric and Luka Sustersic, Moving

pointlike charges and electric and magnetic dipoles, American Journal of Physics 60, 513–519 (1992).

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APPENDIX: Proof of a determinantal formula

Theorem: Let A be an n× n matrix, whose matrix elements are given by:

Aij = δij − aiaj . (89)

Then,

det A = 1− |~a|2 , where |~a|2 =n∑

i=1

a2i . (90)

Proof: The determinant of A is defined as

det A = ǫi1i2···in A1i1A2i2 · · ·Anin ,

where there is an implicit sum over repeated indices. Plugging in eq. (89),

det A = ǫi1i2···iN (δ1i1 − a1ai1)(δ2i2 − a2ai2) · · · (δnin − anain) .

Expanding the product above, and using the Kronecker deltas to perform the sums, we obtain

det A = ǫ123···n − a1ai1ǫi123···n − a2ai2ǫ1i23···n − · · · − anainǫ123···in

= ǫ123···n(1− a21 − a22 − · · · − a2n) , (91)

after performing the final set of summations and using the fact that ǫi1i2···in vanishes unless allof its indices are distinct. Note the absence of any terms in eq. (91) that are quartic (or higherorder) in the ai, since all such terms will be symmetric under the interchange of two indicesthat are summed against two corresponding indices of the Levi-Civita tensor. For example,

ǫi1i2···iNa1ai1a2ai2 = 0 ,

since ai1ai2 is symmetric under the interchange of i1 and i2 whereas ǫi1i2···iN is antisymmetricunder this interchange of indices.

Since ǫ123···n = 1, eq. (91) yields

det A = 1−n∑

i=1

a2i ,

which completes the proof.

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